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I have a Math problem which I did not manage to explain to my IB student boy. $ g(x)= x^3+3x-6 $ admits a single zero at $\alpha \in [1.2, 1.3]$

$f(x)=\frac{x^3+x^2+4}{x^2+1}$ has a derivative which can be written in function of the previous function $g$

$f'(x)= \frac{xg(x)}{(x^2+1)^2}$

Now the question asks to demonstrate that: (1) $f(\alpha)= 1+\frac{3}{2}\alpha$ And this is done by expressing $ f(x) \text{ in function of } g(x)$ then make the polynomial division before equating the whole expression with a fraction $\large \frac{a\color{red}\alpha}{b}$ which gives a=3 and b=2

The Question now asks for the exact value of $f(\alpha)$. I managed already to get the value of $\alpha$ throught Newton-raphson method. I even ploted the function on GeoGebra and went step by step over few elements of the series $x_{i+1}=x_i -\frac{f(x_i)}{f'(x_i)}$ But did managed to get algebraically the value of $f(\alpha)$ without using N-R method.

SAM.Am
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    $f(\alpha)=1+\frac{3}{2}\alpha$ only happen for that particular $\alpha$. So cannot imply $f'(x)=\frac{3}{2}$ or $f'(\alpha)=\frac{3}{2}$. So there is no contradiction. – Abel Wong Oct 12 '22 at 07:55
  • You changed the question a little bit. What is the new question now? To find $\alpha$ algebraically? – Abel Wong Oct 12 '22 at 08:10
  • I figured it out as it's just a single value and not an expression to derive. Sorry, it took me less than a minute to change the question. I meant by algebraically that I woudl like to get the value of $f(\alpha)$ without plotting the function and without using Newton-raphson methode to find $\alpha$ then plugs it. – SAM.Am Oct 12 '22 at 08:22
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    wolframalpha show $\alpha$ is in form of a quite complex surd form. So does $f(\alpha)=1+\frac{3}{2}\alpha$. So if you do not want to use numerical method, you may need to apply the formula of cubic equation solution. link. I guess the question want you to use numerical method. – Abel Wong Oct 12 '22 at 08:35

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